Transcript 4.7 Triangles and Coordinate Proof

4.7 Triangles and
Coordinate Proof
Objectives:
1.
2.
3.
Place geometric figures in a coordinate
plane.
Write a coordinate proof.
Assignment: 2-26 even
Placing Figures in a
Coordinate Plane

So far, you have studied two-column proofs,
paragraph proofs, and flow proofs. A
COORDINATE PROOF involves placing
geometric figures in a coordinate plane.
Then you can use the Distance Formula (no,
you never get away from using this) and the
Midpoint Formula, as well as postulate and
theorems to prove statements about figures.
Ex. 1: Placing a Rectangle in a
Coordinate Plane


Place a 2-unit by 6-unit rectangle in a
coordinate plane.
SOLUTION: Choose a placement that
makes finding distance easy (along the
origin).
Ex. 1: Placing a Rectangle in a
Coordinate Plane

One vertex is at the
origin, and three of the
vertices have at least
one coordinate that is
0.
6
4
2
-5
5
-2
-4
Ex. 1: Placing a Rectangle in a
Coordinate Plane

One side is centered at
the origin, and the xcoordinates are
opposites.
4
2
-5
5
-2
-4
-6
Note:

Once a figure has been placed in a
coordinate plane, you can use the Distance
Formula or the Midpoint Formula to measure
distances or locate points
Ex. 2: Using the Distance
Formula

A right triangle has legs
of 5 units and 12 units.
Place the triangle in a
coordinate plane.
Label the coordinates
of the vertices and find
the length of the
hypotenuse.
6
4
2
5
-2
-4
-6
-8
10
Ex. 2: Using the Distance
Formula
One possible placement is
shown. Notice that one leg
is vertical and the other leg
is horizontal, which assures
that the legs meet as right
angles. Points on the same
vertical segment have the
same x-coordinate, and
points on the same
horizontal segment have
the same y-coordinate.
6
4
2
5
-2
-4
-6
-8
10
Ex. 2: Using the Distance
Formula
You can use the Distance
Formula to find the
length of the
hypotenuse.
d = √(x2 – x1)2 + (y2 – y1)2
= √(12-0)2 + (5-0)2
= √169
= 13
6
4
2
5
-2
-4
-6
-8
10
Ex. 3 Using the Midpoint
Formula


In the diagram, ∆MLO ≅
∆KLO). Find the
coordinates of point L.
Solution: Because the
triangles are congruent, it
follows that ML ≅ KL. So,
point L must be the
midpoint of MK. This
means you can use the
Midpoint Formula to find the
coordinates of point L.
160
140
120
100
80
60
40
20
-50
50
-20
-40
-60
-80
-100
-120
100
150
200
Ex. 3 Using the Midpoint
Formula

160
L (x, y) = x1 + x2, y1 +y2
2
2
140
120
100
Midpoint Formula
80
=160+0 , 0+160
2
2
60
40
20
Substitute values
= (80, 80)
Simplify.
-50
50
-20
-40
-60
-80
-100
-120
100
150
200
Writing Coordinate Proofs

Once a figure is placed in a coordinate plane,
you may be able to prove statements about
the figure.
Ex. 4: Writing a Plan for a
Coordinate Proof
7


Given: Coordinates of
vertices of ∆POS and
∆ROS.
Prove SO bisects
PSR.
6
5
4
S
3
2
1
-5
P
Q
-1
-2
-3
-4
-5
R
5
Ex. 4: Writing a Plan for a
Coordinate Proof




Write a plan to prove that SO bisects PSR.
Given: Coordinates of vertices of ∆POS and ∆ROS.
Prove SO bisects PSR.
Plan for proof: Use the Distance Formula to find the
side lengths of ∆POS and ∆ROS. Then use the
SSS Congruence Postulate to show that ∆POS ≅
∆ROS. Finally, use the fact that corresponding parts
of congruent triangles are congruent (CPOCTAC) to
conclude that PSO ≅RSO, which implies that SO
bisects PSR.
NOTE:


The coordinate proof in Example 4 applies to
a specific triangle. When you want to prove a
statement about a more general set of
figures, it is helpful to use variables as
coordinates.
For instance, you can use variable
coordinates to duplicate the proof in Example
4. Once this is done, you can conclude that
SO bisects PSR for any triangle whose
coordinates fit the given pattern.
No coordinates – just variables
y
S
x
P
(-h, 0)
(0, k)
(0, 0)
R
(h, 0)
Ex. 5: Using Variables as
Coordinates


Right ∆QBC has leg lengths of
h units and k units. You can
find the coordinates of points B
and C by considering how the
triangle is placed in a
coordinate plane.
Point B is h units horizontally
from the origin (0, 0), so its
coordinates are (h, 0). Point C
is h units horizontally from the
origin and k units vertically
from the origin, so its
coordinates are (h, k). You
can use the Distance Formula
to find the length of the
hypotenuse QC.
C (h, k)
hypotenuse
k units
Q (0, 0)
h units
B (h, 0)
Ex. 5: Using Variables as
Coordinates
OC = √(x2 – x1)2 + (y2 – y1)2
= √(h-0)2 + (k - 0)2
= √h2 + k2
C (h, k)
hypotenuse
k units
Q (0, 0)
h units
B (h, 0)
Ex. 5 Writing a Coordinate
Proof

Given: Coordinates of
figure OTUV
Prove ∆OUT  ∆UVO
Coordinate proof:
Segments OV and UT
have the same length.
OV = √(h-0)2 + (0 - 0)2=h

UT = √(m+h-m)2 + (k - k)2=h



6
4
T (m , k)
U (m +h, k)
2
O (0, 0)
-5
-2
-4
-6
V (h,5 0)
Ex. 5 Writing a Coordinate
Proof

Horizontal segments UT
and OV each have a slope
of 0, which implies they are
parallel. Segment OU
intersects UT and OV to
form congruent alternate
interior angles TUO and
VOU. Because OU  OU,
you can apply the SAS
Congruence Postulate to
conclude that ∆OUT 
∆UVO.
6
4
T (m , k)
U (m +h, k)
2
O (0, 0)
-5
-2
-4
-6
V (h,5 0)